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Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a,b) : |a-b| is even} is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

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Vishal Baghel | Contributor-Level 10

5 months ago

We have,

R= ${(a, b) ∴ | a - b |$ is even $}$ is a relation in set A= ${1, 2, 3, 4, 5}$

For all $a \in A$ , $| a - a | = 0$ is even.

So, $(a, a) \in R$ . Hence R is reflexive

For $a, b \in A$ and $(a, b) \in R$

$| a - b |$ is even

$| - b + a |$ is even $|$

$| - (b - a) |$ is even

$| b - a |$ is even

i.e., $(b, a) \in R$

Hence, R is symmetric.

For $a, b, c \in A$ and $(a, b) \in R$ and $(b, c) \in R$

We have $| a - b |$ is even

and $| b - c |$ is even

then, $| a - b | + | b - c |$ is even as even + even=even

$| a - b + b - c |$ is even

$| a - c |$ is even

$∴$ $(a, c) \in R$

So, R is transitive.

$∴$ R is an equivalence relation

All elements of [1,3,5] are odd positive numbers and its subset are odd and their difference given an even number. Hence, they are related to each other.

Similarly,

...more

We have,

R= ${(a, b) ∴ | a - b |$ is even $}$ is a relation in set A= ${1, 2, 3, 4, 5}$

For all $a \in A$ , $| a - a | = 0$ is even.

So, $(a, a) \in R$ . Hence R is reflexive

For $a, b \in A$ and $(a, b) \in R$

$| a - b |$ is even

$| - b + a |$ is even $|$

$| - (b - a) |$ is even

$| b - a |$ is even

i.e., $(b, a) \in R$

Hence, R is symmetric.

For $a, b, c \in A$ and $(a, b) \in R$ and $(b, c) \in R$

We have $| a - b |$ is even

and $| b - c |$ is even

then, $| a - b | + | b - c |$ is even as even + even=even

$| a - b + b - c |$ is even

$| a - c |$ is even

$∴$ $(a, c) \in R$

So, R is transitive.

$∴$ R is an equivalence relation

All elements of [1,3,5] are odd positive numbers and its subset are odd and their difference given an even number. Hence, they are related to each other.

Similarly, all elements of [2,4] are even positive numbers and its subset are even and their difference gives an even number. Hence, they are related to each other.

However, elements of [1,3,5] are related to elements of [2,4] since the difference of the two subsets is never even.

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Let S = {1, 2, 3 ,....,20}

R₁ = {(a, b) : a divide b}

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R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
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$f (x) = {\begin{matrix} e^{- x}, & x < 0 \\ l n x, & x > 0 \end{matrix}$

$g (x) = {\begin{matrix} e^{x}, & x < 0 \\ x, & x > 0 \end{matrix}$

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$= {\begin{array}{l} e^{l n x} = x & (0, 1) & e^{- x} \\ (- \infty, 0) & l n x & (1, \infty) \end{array}$

If A = {1, 2, 3, … 100}, R = {(x, y) | 2x = 3y, x, y ∈ A} is symmetric relation on A and the number of elements in R is n, the smallest integer value of n is

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$?$ R is symmetric relation

⇒ (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

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f is increasing function

x < 5x < 7x

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So total no. of functions = 10⁵ × 1 = 10⁵

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